\chapter{欧拉《无穷小分析引论》十大著名证明}

\author{Euler }
\date{2025.08.27}
	
	% 证明1：欧拉公式
	\section{欧拉公式 $e^{ix} = \cos x + i\sin x$}
	
	\begin{theorem}
		对于任意实数 $x$，有
		\[
		e^{ix} = \cos x + i\sin x.
		\]
		特别地，当 $x = \pi$ 时，有欧拉恒等式 $e^{i\pi} + 1 = 0$。
	\end{theorem}
	
	\begin{proof}
		考虑三个函数的麦克劳林级数展开：
		\[
		\begin{aligned}
			e^{z} &= 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \cdots, \\
			\cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots, \\
			\sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots.
		\end{aligned}
		\]
		令 $z = ix$，代入 $e^z$ 的展开式：
		\[
		\begin{aligned}
			e^{ix} &= 1 + (ix) + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \cdots \\
			&= 1 + ix - \frac{x^2}{2!} - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \cdots \\
			&= \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \right) + i\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \right) \\
			&= \cos x + i\sin x.
		\end{aligned}
		\]
		证毕。
	\end{proof}
	
	\begin{figure}[h]
		\centering
		\begin{tikzpicture}[scale=1.5]
			\draw[->] (-1.5,0) -- (2.5,0) node[right] {$\mathbb{R}$};
			\draw[->] (0,-1.5) -- (0,1.5) node[above] {$\mathbb{I}$};
			\draw (0,0) circle (1cm);
			\def\angle{60}
			\draw[thick, red, ->] (0,0) -- (cos \angle, sin \angle) node[midway, above right] {$r=1$};
			\filldraw[blue] (cos \angle, sin \angle) circle (1pt) node[right] {$e^{ix} = \cos x + i\sin x$};
			\draw (0.3,0) arc (0:\angle:0.3) node[midway, right] {$x$};
			\draw[dashed] (cos \angle, sin \angle) -- (cos \angle, 0) node[below] {$\cos x$};
			\draw[dashed] (cos \angle, sin \angle) -- (0, sin \angle) node[left] {$\sin x$};
		\end{tikzpicture}
		\caption{欧拉公式的几何意义}
	\end{figure}
	
	% 证明2：正弦和余弦的级数展开
	\section{正弦和余弦函数的级数展开}
	
	\begin{theorem}
		\[
		\begin{aligned}
			\sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \\
			\cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots
		\end{aligned}
		\]
	\end{theorem}
	
	\begin{proof}
		由欧拉公式 $e^{ix} = \cos x + i\sin x$ 和 $e^{ix} = \sum_{n=0}^{\infty} \frac{(ix)^n}{n!}$，
		比较实部和虚部即得证。
	\end{proof}
	
	\begin{figure}[h]
		\centering
		\begin{tikzpicture}[declare function = {
				sin1(\x) = \x;
				sin2(\x) = \x - (\x^3)/6;
				sin3(\x) = \x - (\x^3)/6 + (\x^5)/120;
				sinreal(\x) = sin(\x r);}]
			\begin{axis}[
				xmin = -7, xmax = 7, ymin = -2, ymax = 2,
				axis lines = middle, xlabel = $x$, ylabel = $y$,
				legend pos = south west]
				\addplot[domain=-6.5:6.5, samples=100, thick, blue] {sinreal(x)};
				\addplot[domain=-6.5:6.5, samples=100, red, dashed] {sin1(x)};
				\addplot[domain=-6.5:6.5, samples=100, green, dashed] {sin2(x)};
				\addplot[domain=-6.5:6.5, samples=100, orange, dashed] {sin3(x)};
				\legend{$\sin x$, 1项近似, 2项近似, 3项近似}
			\end{axis}
		\end{tikzpicture}
		\caption{正弦函数及其泰勒级数逼近}
	\end{figure}
	
	% 证明3：巴塞尔问题
	\section{巴塞尔问题：$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$}
	
	\begin{theorem}
		\[
		\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.
		\]
	\end{theorem}
	
	\begin{proof}
		考虑 $\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots$ 和
		$\frac{\sin x}{x} = \prod_{n=1}^{\infty} \left(1 - \frac{x^2}{n^2 \pi^2}\right)$。
		比较 $x^2$ 项系数：$-\frac{1}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n^2} = -\frac{1}{6}$，
		故 $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$。
	\end{proof}
	
	\begin{figure}[h]
		\centering
		\begin{tikzpicture}
			\begin{axis}[
				xmin=0, xmax=7, ymin=0, ymax=1.2,
				axis lines=middle, xlabel=$n$, ylabel=$y$,
				xtick={1,2,...,6}, ytick={0,0.5,1},
				legend pos=north east]
				\addplot+[ycomb, blue, thick] coordinates {
					(1, 1) (2, 1/4) (3, 1/9) (4, 1/16) (5, 1/25) (6, 1/36)};
				\addplot[domain=0.7:6.3, red, thick] {1/x^2};
				\legend{项: $1/n^2$, 曲线: $y=1/x^2$}
			\end{axis}
		\end{tikzpicture}
		\caption{巴塞尔问题级数项图示}
	\end{figure}
	
	% 证明4：指数函数 e 的定义
	\section{指数函数 $e$ 的定义和性质}
	
	\begin{definition}[自然指数底 $e$]
		\[
		e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{\infty} \frac{1}{k!}.
		\]
	\end{definition}
	
	\begin{proof}
		二项式展开：
		\[
		\left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k}
		= 1 + 1 + \frac{1}{2!}\left(1-\frac{1}{n}\right) + \cdots
		\]
		当 $n \to \infty$ 时，趋于 $\sum_{k=0}^{\infty} \frac{1}{k!}$。
	\end{proof}
	
	\begin{figure}[h]
		\centering
		\begin{tikzpicture}
			\begin{axis}[
				xmin=0, xmax=20, ymin=2, ymax=3,
				axis lines=middle, xlabel=$n$, ylabel=$y$,
				legend pos=south east]
				\addplot[blue, thick, domain=1:20, samples=20] {(1+1/x)^x};
				\addplot[red, dashed] {2.71828};
				\legend{$(1+1/n)^n$, $y = e$}
			\end{axis}
		\end{tikzpicture}
		\caption{序列 $(1+1/n)^n$ 收敛于 $e$}
	\end{figure}
	
	% 证明5：Gamma函数
	\section{Gamma函数：阶乘的推广}
	
	\begin{definition}[Gamma函数]
		\[
		\Gamma(z) = \int_0^{\infty} t^{z-1} e^{-t} dt, \quad \operatorname{Re}(z) > 0.
		\]
	\end{definition}
	
	\begin{theorem}
		对于正整数 $n$，有 $\Gamma(n) = (n-1)!$。
	\end{theorem}
	
	\begin{proof}
		分部积分：$\Gamma(z+1) = z\Gamma(z)$，且 $\Gamma(1) = 1$，
		故 $\Gamma(n) = (n-1)\Gamma(n-1) = (n-1)!$。
	\end{proof}
	
	\begin{figure}[h]
		\centering
		\begin{tikzpicture}
			\begin{axis}[
				xmin=0, xmax=5, ymin=0, ymax=25,
				axis lines=middle, xlabel=$x$, ylabel=$\Gamma(x)$,
				samples=50]
				\addplot[blue, thick, domain=0.1:4.9] {gamma(x)};
				\addplot[red, only marks, mark=*] coordinates {
					(1, 1) (2, 1) (3, 2) (4, 6) (5, 24)};
				\legend{$\Gamma(x)$, $(n-1)!$}
			\end{axis}
		\end{tikzpicture}
		\caption{Gamma函数与阶乘的关系}
	\end{figure}
	
	% 证明6：圆的面积公式
	\section{圆的面积与周长公式}
	
	\begin{theorem}
		半径为 $r$ 的圆，$C = 2\pi r$，$A = \pi r^2$。
	\end{theorem}
	
	\begin{proof}
		将圆分为无穷多个扇形，每个面积 $\frac{1}{2} r^2 d\theta$，
		积分得 $A = \int_0^{2\pi} \frac{1}{2} r^2 d\theta = \pi r^2$。
	\end{proof}
	
	\begin{figure}[h]
		\centering
		\begin{tikzpicture}[scale=1.5]
			\draw (0,0) circle (1.5cm);
			\filldraw[fill=blue!20, draw=blue] (0,0) -- (0.3,0) arc (0:30:0.3) -- cycle;
			\draw (0,0) -- (1.5, 0) node[midway, below] {$r$};
			\draw (0,0) -- (30:1.5) node[midway, above] {$r$};
			\draw (0.5,0) arc (0:30:0.5) node[right] {$d\theta$};
		\end{tikzpicture}
		\caption{圆的面积推导图示}
	\end{figure}
	
	% 证明7：正弦函数的无穷乘积
	\section{正弦函数的无穷乘积公式}
	
	\begin{theorem}
		\[
		\sin x = x \prod_{n=1}^{\infty} \left(1 - \frac{x^2}{n^2 \pi^2}\right).
		\]
	\end{theorem}
	
	\begin{proof}
		将 $\frac{\sin x}{x}$ 视为具有零点 $x = \pm n\pi$ 的整函数，应用因子定理。
	\end{proof}
	
	\begin{figure}[h]
		\centering
		\begin{tikzpicture}
			\begin{axis}[
				xmin=-4, xmax=4, ymin=-1.5, ymax=1.5,
				axis lines=middle, xlabel=$x$, ylabel=$y$,
				samples=100]
				\addplot[blue, thick, domain=-4:4] {sin(deg(x))};
				\addplot[red, thick, domain=-4:4] {x};
				\addplot[green, thick, domain=-4:4] {x*(1-x^2/(pi^2))};
				\addplot[orange, thick, domain=-4:4] {x*(1-x^2/(pi^2))*(1-x^2/(4*pi^2))};
				\legend{$\sin x$, 1项乘积, 2项乘积, 3项乘积}
			\end{axis}
		\end{tikzpicture}
		\caption{正弦函数无穷乘积逼近}
	\end{figure}
	
	% 证明8：二项式定理推广
	\section{二项式定理的推广（分数指数）}
	
	\begin{theorem}
		对于任意实数 $\alpha$，有：
		\[
		(1 + x)^\alpha = \sum_{k=0}^{\infty} \binom{\alpha}{k} x^k, \quad |x| < 1.
		\]
	\end{theorem}
	
	\begin{proof}
		考虑 $(1+x)^\alpha$ 的泰勒级数展开并计算各阶导数。
	\end{proof}
	
	\begin{figure}[h]
		\centering
		\begin{tikzpicture}
			\begin{axis}[
				xmin=-1, xmax=1, ymin=0, ymax=2,
				axis lines=middle, xlabel=$x$, ylabel=$y$,
				samples=100]
				\addplot[blue, thick, domain=-0.99:1] {(1+x)^0.5};
				\addplot[red, dashed, domain=-0.8:0.8] {1 + 0.5*x};
				\addplot[green, dashed, domain=-0.8:0.8] {1 + 0.5*x - 0.125*x^2};
				\addplot[orange, dashed, domain=-0.8:0.8] {1 + 0.5*x - 0.125*x^2 + 0.0625*x^3};
				\legend{$(1+x)^{0.5}$, 1项近似, 2项近似, 3项近似}
			\end{axis}
		\end{tikzpicture}
		\caption{二项式定理推广图示}
	\end{figure}
	
	% 证明9：正切函数连分式
	\section{正切函数的连分式展开}
	
	\begin{theorem}
		\[
		\tan x = \cfrac{x}{1 - \cfrac{x^2}{3 - \cfrac{x^2}{5 - \cfrac{x^2}{7 - \cdots}}}}.
		\]
	\end{theorem}
	
	\begin{proof}
		欧拉通过函数方程推导，验证可通过与泰勒级数比较。
	\end{proof}
	
	\begin{figure}[h]
		\centering
		\begin{tikzpicture}
			\begin{axis}[
				xmin=-1.5, xmax=1.5, ymin=-5, ymax=5,
				axis lines=middle, xlabel=$x$, ylabel=$y$,
				samples=100, restrict y to domain=-10:10]
				\addplot[blue, thick, domain=-1.4:1.4] {tan(deg(x))};
				\addplot[red, dashed, domain=-0.5:0.5] {x};
				\addplot[green, dashed, domain=-0.8:0.8] {x/(1-x^2/3)};
				\legend{$\tan x$, 1项近似, 2项近似}
			\end{axis}
		\end{tikzpicture}
		\caption{正切函数连分式逼近}
	\end{figure}
	
	% 证明10：傅里叶级数前身
	\section{傅里叶级数的前身：三角级数展开}
	
	\begin{theorem}
		周期函数可表示为：
		\[
		f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos nx + b_n \sin nx \right).
		\]
	\end{theorem}
	
	\begin{proof}
		欧拉推导系数公式，如 $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx dx$，
		利用三角函数的正交性。
	\end{proof}
	
	\begin{figure}[h]
		\centering
		\begin{tikzpicture}
			\begin{axis}[
				xmin=-3.5, xmax=3.5, ymin=-1.5, ymax=1.5,
				axis lines=middle, xlabel=$x$, ylabel=$y$,
				samples=100]
				\addplot[blue, thick, domain=-3.14:3.14] {x/3.14};
				\addplot[red, thick, domain=-3.14:3.14] {2/3.14 * sin(deg(x))};
				\addplot[green, thick, domain=-3.14:3.14] {2/3.14 * (sin(deg(x)) - sin(deg(2*x))/2)};
				\legend{$f(x)=x/\pi$, 1项逼近, 2项逼近}
			\end{axis}
		\end{tikzpicture}
		\caption{傅里叶级数逼近图示}
	\end{figure}
	